Question The inductor shown in Figure 1 consists of two E shape cores with a gap and a coil in the centre leg. The relative permeability of the magnetic material is 1500. (a) Draw the equivalent magnetic circuit and calculate the equivalent reluctance of the structure; [8 marks] (b) Calculate the self-inductance for N=35 turns; [3 marks] (c) Calculate the energy stored in the inductor for a coil current such that the flux in the centre limb is 116 µWb. 34 . Ꮎ ᎾᎿ 34 0.5 Ꭲ 26 26 8 10 8 4 10 Figure 1 Core Dimensions in mm
The inductor shown in Figure 1 consists of two E shape cores with a gap and a coil in the centre leg. The relative permeability of the magnetic material is 1500.
(a) Draw the equivalent magnetic circuit and calculate the equivalent reluctance of the structure; [8 marks]
(b) Calculate the self-inductance for N=35 turns; [3 marks]
(c) Calculate the energy stored in the inductor for a coil current such that the flux in the centre limb is 116 µWb.
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(a)\mu_{n}=500Area of Lef limb = Anea of right tiun=10 \times 4=40 \times 10^{-9} \mathrm{~m}\begin{array}{l}R_{1} \text { - recuetance of lef is } \\R_{R}-\cdots \text { Right } \\R_{g}=\cdots \text { Air gap } \\R_{L c}-1.8 \text { lowacenera } \\\end{array}N=35R_{u c}-1 \text { \&uppes limes }\begin{aligned}\text { mimf } & =N \underline{C} \\\varnothing & =116 \times 10^{-6} \mathrm{Nb}\end{aligned}\varnothing=116 \times 10^{-6} \mathrm{Nb}I=\text { ? }\begin{array}{l}R_{L}=\frac{l_{L}}{\mu_{0} \mu_{1} A} \\I_{L}=44-4+\frac{34}{}-4 \Rightarrow 60 \mathrm{~mm} \\I_{R}=\eta_{L}=60 \times 10^{-3} \mathrm{~m}=60 \times 10^{-3 \mathrm{~m}} \\=\frac{60 \times 10^{-3}}{4 \pi \times 10^{-7} \times 500 \times 40 \times 10^{-6}} \\\eta_{g}=0.5 \mathrm{~mm} \Rightarrow 0.5 \times 10^{-3} \mathrm{~m} \\I_{k}=q_{v c}=\frac{34}{2}-2-\frac{0.5}{2} \Rightarrow 14.75 \mathrm{~mm} \\=14.75 \times 10^{-3} \mathrm{~m} \\=2.387 \times 10^{6} \mathrm{AT} / \mathrm{Wb} \\\end{array}Area ficentes limb\begin{array}{l}R_{L}=R_{R}=2.387 \times 10^{6} \mathrm{AT} / \omega \mathrm{b} \\A_{c}=10 \times 10=100 \times 10^{-6} \mathrm{~m}^{2} \\R_{g}=\frac{9_{9}}{\mu_{0} A_{C}} \Rightarrow \frac{0.5 \times 10^{-3}}{4 \pi \times 10^{-7} \times 100 \times 10^{-6}} \Rightarrow 3.97 \times 10^{6} \mathrm{\pi} \% \mathrm{ub} \\R_{U C}=R_{L C}=\frac{R_{C C}}{\mu_{0} \mu_{A} A_{C}} \Rightarrow \frac{14.75 \times 10^{-3}}{47 \times 10^{-7} \times 100 \times 10^{-6} \mathrm{Ar} / \mathrm{bb}} \\\end{array}R_{e q}=R_{L}\left\{_{1} R_{x}=R_{g}+R_{u L}+R_{L} c\right.\begin{aligned}R_{x} & =R_{g}+R_{U L}+R_{L C} \\& =238.72 \mathrm{M} \frac{A T}{C_{2 L}}\end{aligned}R_{L} \text { is paralle with } R_{R} \Rightarrow \frac{R_{L} \times R_{n}}{R_{L}+R_{n}} \Rightarrow 1.19 M \frac{A T}{\Delta b}<R_{y}\begin{aligned}R_{e q} & =R_{u}+R_{y} \\& =1.19 \mathrm{M}+238.72 \mathrm{M} \\& =240 \mathrm{MA7} / \mathrm{Lb}\end{aligned}\begin{aligned}\min =N ! & =\varnothing \times R_{e q} \\? & =\frac{\varnothing R_{e q}}{\mathbb{N}} \Rightarrow \frac{116 \times 10^{-6} \times 240 \times 10^{6}}{35} \\& =795.4 \mathrm{Amp}\end{aligned}(ii)Self Inductence, L=\frac{N \varphi}{2} \Rightarrow \frac{35 \times 116 \times 10^{-6}}{795.4}=5.1 \mathrm{MH}(iii)Cvergy stored,\begin{aligned}E & =\frac{1}{2} L I^{2} \\& =\frac{1}{2} \times 5.1 \times 10^{-6} \times 795.4^{2} \\& =1.61 \text { joules }\end{aligned} ...